Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include"bits/stdc++.h"
#define int long long
#define REP(i,n) for(int i=0,i##_len=(n);i<i##_len;++i)
using namespace std;
const int MAX_N=2999;
int N;
vector<double> p;
double dp[MAX_N+1][MAX_N+1];
signed main(){
	cin>>N;
	p.resize(N);
	REP(i,N) cin>>p[i];
	REP(i,(N+1)/2) dp[N-i][i]=1;
	for(int i=N-1;i>=0;--i){
		for(int j=N-1;j>=0;--j){
			if(i+j<N) dp[i][j]=dp[i+1][j]*p[i+j]+dp[i][j+1]*(1-p[i+j]);
		}
	}
	printf("%.16f\n",dp[0][0]);
}